Practice problems for exam

Problem 1

A = [2 -2 1; 3 2 -3; 1 1 5]; A = A([3 2 1],:)
A = 3×3
     1     1     5
     3     2    -3
     2    -2     1
[L,U,p] = lu(A,'vector')
L = 3×3
    1.0000         0         0
    0.6667    1.0000         0
    0.3333   -0.1000    1.0000
U = 3×3
    3.0000    2.0000   -3.0000
         0   -3.3333    3.0000
         0         0    6.3000
p = 1×3
     2     3     1

Problem 2

L = [1 0 0; -2 1 0; 2 -3 1]; 
U = [2 1 -1; 0 -3 2; 0 0 1];
p = [3; 1; 2];
b = [2; 1; 4];
y = L\b(p)    % solve with forward substitution
y = 3×1
     4
    10
    23
x = U\y       % solve with back substitution
x = 3×1
    7.5000
   12.0000
   23.0000

Problem 3

A = [-7 8 -8; -8 9 -9; 6 -7 8];
b = [.97; 1.02; -.01];
xhat = [1; 2; 1];
bhat = A*xhat
bhat = 3×1
     1
     1
     0
epsb = norm(bhat-b,1)/norm(bhat,1)
epsb = 0.0300

Solution 1: We use

with

normA = norm(A,1)
normA = 25
bound = norm(A,1)*25*epsb
bound = 18.7500

Solution 2: We use

, hence

and

Therefore

bound = 25*norm(bhat-b,1)/norm(xhat,1)
bound = 0.3750

This is the error bound

, see "Errors for linear systems".

Problem 4

t = [0; 1; 3];
y = [2; 4; 3]
y = 3×1
     2
     4
     3
A = [t, t.^2]
A = 3×2
     0     0
     1     1
     3     9
M = A'*A
M = 2×2
    10    28
    28    82
v = A'*y
v = 2×1
    13
    31
c = M\v    % solve normal equations A*c = v
c = 2×1
    5.5000
   -1.5000

Instead of the normal equations we can also use the Matlab \ command:

c = A\y
c = 2×1
    5.5000
   -1.5000

Problem 5

f = @(x) x^3+x-11;

(a) Secant method

x0 = 3; x1 = 2;
x2 = x1 - f(x1)*(x1-x0)/(f(x1)-f(x0))
x2 = 2.0500

(b) Newton method

fp = @(x) 3*x^2 + 1;    % f'(x)
x0 = 2;
x1  = x0 - f(x0)/fp(x0)
x1 = 2.0769

and

have different signs. Hence there is a solution in the interval

. (Since

the function f is strictly increasing, and the solution is unique.)

xs = fzero(f,[2,3])
xs = 2.0743

Problem 6

Define the function

f = @(x) [4*x(1)+x(1)^2*x(2)-4; 4*x(2)-x(1)*x(2)^2-3];

Find the Jacobian

fp = @(x) [4 + 2*x(1)*x(2), x(1)^2; -x(2)^2, 4 - 2*x(1)*x(2)];

Perform one step of the Newton method starting with

x0 = [1;0];
b = f(x0)
b = 2×1
     0
    -3
A = fp(x0)
A = 2×2
     4     1
     0     4
d = -A\b        % solve linear system A*d = -b
d = 2×1
   -0.1875
    0.7500
x1 = x0 + d
x1 = 2×1
    0.8125
    0.7500

Iterate Newton method until

x = [1;0]
x = 2×1
     1
     0
while 1
  b = f(x);
  if norm(b,1)<=1e-10
    break
  end
  A = fp(x);
  d = -A\b;
  x = x + d;
end
x
x = 2×1
    0.8369
    0.9316
normf = norm(b,1)
normf = 8.8818e-16

Use fsolve (with default settings):

xs = fsolve(f,[1;0])
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

<stopping criteria details>
xs = 2×1
    0.8369
    0.9316
norm(f(xs),1)
ans = 9.6660e-10